HDU 1074 Doing Homework（状压DP）

第一次写博客ORZ……

http://acm.split.hdu.edu.cn/showproblem.php?pid=1074

http://acm.hdu.edu.cn/showproblem.php?pid=1074

这两个总有一个是可以点开的……

题意比较清晰的啦。

做法的话暴力显然不合适，15！太大。

所以考虑状压DP……

代码参考了网上大神的。十分感谢。谢谢@键盘上的舞者

http://blog.csdn.net/libin56842/article/details/24316493

细节问题比较麻烦。在扣分相同的情况下要按字母输出课程名称。这个地方我卡了半天呢。

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>

using namespace std;

#define REP(i,n)        for(int i(0); i <  (n); ++i)
#define rep(i,a,b)        for(int i(a); i <= (b); ++i)
#define dec(i,a,b)        for(int i(a); i >= (b); --i)
#define for_edge(i,x)        for(int i = H[x]; i; i = X[i])

#define LL    long long
#define ULL    unsigned long long
#define MP    make_pair
#define PB    push_back
#define FI    first
#define SE    second
#define INF    1 << 26

const int N    =    100000        +    10;
const int M    =    10000        +    10;
const int Q    =    1000        +    10;
const int A    =    15        +    1;

struct node{
    int time, score, pos, pre;
} dp[1 << A];

int cost[A];
char name[A][Q];
int dead[Q];
int T;
int n;


void print(int k){
    if (dp[k].pre) print(dp[k].pre);
    puts(name[dp[k].pos]);
}    
    
int main(){
#ifndef ONLINE_JUDGE
    freopen("test.txt", "r", stdin);
    freopen("test.out", "w", stdout);
#endif

    scanf("%d", &T);
    REP(Case, T){
        memset(dp, 0, sizeof dp);
        scanf("%d", &n);
        REP(i, n) scanf("%s%d%d", name[i], dead + i, cost + i);
        int S = (1 << n) - 1;
        rep(now, 1, S){
            dp[now].score = INF;
            bool flag = false;
            REP(i, n) if (now & (1 << i)){
                int past = now ^ (1 << i);
                int st = dp[past].time + cost[i] - dead[i]; st = max(st, 0);
                if (st + dp[past].score < dp[now].score || (flag && st + dp[past].score == dp[now].score && strcmp(name[dp[now].pos], name[i]) < 0)){
                    flag = true;
                    dp[now].score = st + dp[past].score; dp[now].pos = i;
                    dp[now].pre = past; dp[now].time = dp[past].time + cost[i];
                }
            }
        }
        
        printf("%d\n", dp[S].score);
        print(S);
        
    }
    
    
    
    return 0;
    
}